modeling and simulation of dynamic systems

MODELING AND SIMULATION
OF DYNAMIC SYSTEMS

MIXED DISCIPLINE SYSTEMS

PHAM HUY HOANG
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY

INTRODUCTION

MIXED DISCIPLINE SYSTEM:
MIXED DISCIPLINE SYSTEM – COUPLING SYSTEM OF
SINGLE-DISCIPLINE SYSTEMS

Pham Huy Hoang

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ELECTROMECHANICAL SYSTEMS
ARMATURE-CONTROLLED DC MOTOR
Voltage is electric potential energy per unit charge
(J/C = V) - referred to as "electric potential”.
Electromotive force (emf) voltage (electromotance):
- is that which tends to cause current (actual
electrons and ions) to flow;
- is the external work expended per unit of charge
to produce an electric potential difference across
two open-circuited terminals;
- is generated by a magnetic force (Faraday’s
law).

Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS
Faraday's Law
Any change in the
magnetic
environment* of a coil
of wire will cause a
voltage (emf) to be
"induced" in the coil.
* The change of
magnetic field
strength, relative
displacement
between the magnet
field and the coil.
Pham Huy Hoang

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Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS
The back emf voltage across a DC motor:
eb = K eω = K eθ&

The torque developed by the motor:
T = Kt i

eb : back emf voltage.
θ : angular displacement of the rotor of the motor
&
θ = ω : angular velocity of the rotor
T : torque applied to the rotor
Ke : emf constant (Vs/rad)
Ki : torque constant (Nm/A)
Pham Huy Hoang

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ELECTROMECHANICAL SYSTEMS
ia

Ra

La

θ ,θ& = ω
Jr

eb

Va

TL

Jd

Bd

vRa + vLa + eb − va = 0
di
Raia + La a + eb = va
dt
eb = K eω = K eθ&
Raia + La

dia
+ K eθ& = va
dt

(1)
Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS
ia

Ra

La

θ ,θ& = ω
eb

Va

Jr

TL

Jd

Bd

J = Jr + Jd
T + TL − Bdθ& = Jθ&&
T = Kt ia
Kt ia + TL − Bdθ& = Jθ&& (2)
Pham Huy Hoang

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ELECTROMECHANICAL SYSTEMS
ia

Ra

La

θ ,θ& = ω
eb

Va

Jr

TL

Jd

Bd

Jθ&& + Bdθ& − Kt ia = TL
di
La a + Raia + K eθ& = va
dt
θ&&
θ&
 J 0    Bd 0    0 − Kt  θ  TL 
 0 0  ..  +  K L   .  + 0 R  i  = v 

 i   e
a  i  
a  a   a 
 a
a
Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS

ia

Va

Ra

La

θ ,θ& = ω
eb

Jr
K, B

TL

Jd
Bd

Pham Huy Hoang

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ELECTROMECHANICAL SYSTEMS
FIELD-CONTROLLED DC MOTOR
Rf
θ ,θ& = ω
ia = const
if
Jr
vf
Jd
Lf
eb= 0

Rf if + Lf

Bd

T = Kt i f

di f
dt

= vf

TL

(1)

Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS
Rf
vf

ia = const

if

Jr

Lf
eb= 0

θ ,θ& = ω

T = Kt i f

TL

Jd
Bd

J = Jr + Jd
T + TL − Bdθ& = Jθ&&
T = Kt i f
Kt i f + TL − Bdθ& = Jθ&& (2)
Pham Huy Hoang

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ELECTROMECHANICAL SYSTEMS
Rf
vf

if

Jr

Lf
eb= 0

θ ,θ& = ω

ia = const

TL

Jd
Bd

T = Kt i f

Jθ&& + Bd θ& − K t i f = TL
di f
Lf
+Rf if =vf
dt
&&
0  θ   Bd − K t  θ&  TL 
J
=
 0 L  .  +  0
R f  i f  v f 
f  i  

 f
Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS

Rf
vf

ia = const

if

Jr

Lf
eb= 0

θ ,θ& = ω

T = Kt i f

TL

Jd

Bd

Pham Huy Hoang

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ELECTROMECHANICAL SYSTEMS
MAGNETO-ELECTRO-MECHANICAL SYSTEMS
Lenz’s law: increasing current
in a coil will generate a
counter emf which opposes
the current. (The emf always

opposes the change in
current).

The relation of this counter emf to the current is the
origin of the concept of inductance.
Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS
Magnetic Force and Lorentz force law:
- The force is perpendicular to both the velocity v of
the charge q and the magnetic field B (N/A = Ns/C =
Tesla).
- The magnitude of the force is F = qvB sinθ
(θ is the angle between the velocity and the magnetic
field).

Pham Huy Hoang

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ELECTROMECHANICAL SYSTEMS

-The magnetic force on a stationary charge or a
charge moving parallel to the magnetic field is zero.
- The direction of the force is given by the right hand
rule.

Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS
i1
v

i2

R2

R1

L

C

k1
y

b

m1

a

Magnetic force : K1i2
Electromotive force (emf ) : K 2 y&

c1

k2

x1
c2

m2

x2

Pham Huy Hoang

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ELECTROMECHANICAL SYSTEMS
i1
v

i2
R2

R1

L

C

y

1t
1t
R1i1 + ∫ i1dt − ∫ i2dt = v

C0
C0

a

1t
di2
1t
b
− ∫ i1dt + L
+ R2i2 + ∫ i2dt − K 2 x&1 = 0
C0
dt
C0
a
Pham Huy Hoang

ELECTROMECHANICAL SYSTEMS
∑ M O = J Oα
&x&
− K1i2a − (k1x1 + c1x&1)b + [k2 ( x2 − x1) + c2 ( x&2 − x&1)]b = m1b2  1 
b
m1&x&1 + (c1 + c2 )b2 x&1 − c2b2 x&2 + (k1 + k2 )b2 x1 − k2b2 x2 + K1abi2 = 0
∑ F = m2 &x&2
− k2 ( x2 − x1) − c2 ( x&2 − x&1) = m2 &x&2

k1

m2 &x&2 − c2 x&1 + c2 x&2 − k2 x1 + k2 x2 = 0

b

c1
m1

k2

x1
c2

m2

x2

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